Value Error: too many values to unpack (expected 3)
Value Error: too many values to unpack (expected 3)
我正在尝试实现值迭代算法。
我有一个网格
grid = [[0, 0, 0, +1],
[0, "W", 0, -1],
[0, 0, 0, 0]]
一个动作列表
actlist = {UP:1, DOWN:2, LEFT:3, RIGHT:4}
还有一个奖励函数
reward = [[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0]]
我写了一个函数 T，它 returns 3 个元组的元组。
def T(i,j,actions):
if(i == 0 and j == 0):
if(actions == UP):
return (i,i,0.8),(i,i,0.1),(i,j+1,0.1)
elif(actions == DOWN):
return (i+1,j,0.8),(i,j,0.1),(i,j+1,0.1)
elif(actions == LEFT):
return (i,j,0.8),(i,j,0.1),(i+1,j,0.1)
elif(actions == RIGHT):
return (i,j+1,0.8),(i,i,0.1),(i+1,j,0.1)
elif (i == 0 and j == 1):
if(actions == UP):
return (i,i,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == DOWN):
return (i,j,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == LEFT):
return (i,j-1,0.8),(i,j,0.1),(i,j,0.1)
elif(actions == RIGHT):
return (i,j+1,0.8),(i,j,0.1),(i,j,0.1)
elif(i == 0 and j == 2):
if(actions == UP):
return (i,j,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == DOWN):
return(i+1,j,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == LEFT):
return (i,j-1,0.8),(i,j,0.1),(i+1,j,0.1)
elif(actions == RIGHT):
return (i,j+1,0.8),(i,j,0.1),(i+1,j,0.1)
elif(i == 0 and j == 3):
if(actions == UP):
return (-1,-1,0.8),(-1,-1,0.1),(-1,-1,0.1)
elif(actions == DOWN):
return (-1,-1,0.8),(-1,-1,0.1),(-1,-1,0.1)
elif(actions == LEFT):
return (-1,-1,0.8),(-1,-1,0.1),(-1,-1,0.1)
elif(actions == RIGHT):
return (-1,-1,0.8),(-1,-1,0.1),(-1,-1,0.1)
# 2nd row
elif (i == 1 and j == 0):
if(actions == UP):
return (i-1,j,0.8),(i,j,0.1),(i,j,0.1)
elif(actions == DOWN):
return (i+1,j,0.8),(i,j,0.1),(i,j,0.1)
elif(actions == LEFT):
return (i,j,0.8),(i-1,j,0.1),(i+1,j,0.1)
elif(actions == RIGHT):
return (i,j,0.8),(i-1,j,0.1),(i+1,j,0.1)
elif(i == 1 and j ==1):
if(actions == UP):
return (i,j,0.8),(i,j,0.1),(i,j,0.1)
elif(actions == DOWN):
return (i,j,0.8),(i,j,0.1),(i,j,0.1)
elif(actions == LEFT):
return (i,j,0.8),(i,j,0.1),(i,j,0.1)
elif(actions == RIGHT):
return (i,j,0.8),(i,j,0.1),(i,j,0.1)
elif (i == 1 and j == 2):
if(actions == UP):
return (i-1,j,0.8),(i,j,0.1),(i,j+1,0.1)
elif(actions == DOWN):
return (i+1,j,0.8),(i,j,0.1),(i,j+1,0.1)
elif(actions == LEFT):
return (i,j,0.8),(i-1,j,0.1),(i+1,j,0.1)
elif(actions == RIGHT):
return (i,j+1,0.8),(i-1,j,0.1),(i+1,j,0.1)
elif(i == 1 and j == 3):
if(actions == UP):
return (-2,-2,0.8),(-2,-2,0.1),(-2,-2,0.1)
elif(actions == DOWN):
return (-2,-2,0.8),(-2,-2,0.1),(-2,-2,0.1)
elif(actions == LEFT):
return (-2,-2,0.8),(-2,-2,0.1),(-2,-2,0.1)
elif(actions == RIGHT):
return (-2,-2,0.8),(-2,-2,0.1),(-2,-2,0.1)
# 3rd row
elif(i == 2 and j == 0):
if(actions == UP):
return (i-1,j,0.8),(i,j,0.1),(i,j+1,0.1)
elif(actions == DOWN):
return (i,j,0.8),(i,j,0.1),(i,j+1,1,0.1)
elif(actions == LEFT):
return (i,j,0.8),(i-1,j,0.1),(i,j,0.1)
elif(actions == RIGHT):
return (i,j+1,0.8),(i-1,j,0.1),(i,j,0.1)
elif (i == 2 and j == 1):
if(actions == UP):
return (i,j,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == DOWN):
return (i,j,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == LEFT):
return (i,j-1,0.8),(i,j,0.1),(i,j,0.1)
elif(actions == RIGHT):
return (i,j+1,0.8),(i,j,0.1),(i,j,0.1)
elif(i == 2 and j == 2):
if(actions == UP):
return (i-1,j,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == DOWN):
return (i,j,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == LEFT):
return (i,j-1,0.8),(i-1,j,0.1),(i,j,1)
elif(actions == RIGHT):
return (i,j+1,0.8),(i-1,j,0.1),(i,j,0.1)
elif(i == 2 and j == 3):
if(actions == UP):
return (i-1,j,0.8),(i,j-1,0.1),(i,j,0.1)
elif(actions == DOWN):
return (i,j,0.8),(i,j-1,0.1),(i,j,0.1)
elif(actions == LEFT):
return (i,j-1,0.8),(i-1,j,0.1),(i,j,0.1)
elif(actions == RIGHT):
return (i,j,0.8),(i-1,j,0.1),(i,j,0.1)
这个函数在值迭代函数中被调用：
def value_iteration():
U1 = [[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0]]
while True:
U=U1.copy()
delta = 0
for i in range(len(grid)):
for j in range(len(grid[i])):
U1[i][j] = max(sum(p*(R(k,l)+gamma*U[k][l]) for (k,l,p) in T(i,j,a)) for a in actlist)
print(i,j,U1[i][j])
delta = max(delta, abs(U1[i][j] - U[i][j]))
if delta <= epsilon*(1 - gamma)/gamma:
return U
问题是，for 循环的前两次迭代与输出相得益彰
0 0
0 1
0 2
0 3
1 0
1 1
1 2
1 3
但随后代码因错误停止
ValueError: too many values to unpack (expected 3)
看看** **中的元组，也许就是这个原因
# 3rd row
elif(i == 2 and j == 0):
if(actions == UP):
return (i-1,j,0.8),(i,j,0.1),(i,j+1,0.1)
elif(actions == DOWN):
return (i,j,0.8),(i,j,0.1),**(i,j+1,1,0.1)**
elif(actions == LEFT):
return (i,j,0.8),(i-1,j,0.1),(i,j,0.1)
elif(actions == RIGHT):
return (i,j+1,0.8),(i-1,j,0.1),(i,j,0.1)
elif (i == 2 and j == 1):
if(actions == UP):
return (i,j,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == DOWN):
return (i,j,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == LEFT):
return (i,j-1,0.8),(i,j,0.1),(i,j,0.1)
elif(actions == RIGHT):
return (i,j+1,0.8),(i,j,0.1),(i,j,0.1)
正如@EdwardMinnix 提到的，您应该使用 map 而永远不要进行这种 if/else 构造。或者，如果有任何具有这些值的模式，请查看 Stragety Pattern.
这会对你有所帮助：
...
VALUE_A = 0.8
VALUE_B = 0.1
def new_T(i, j, actions):
result_map = {(0, 0, 1): ((i, i, VALUE_A), (i, i, VALUE_B), (i, j + 1, VALUE_B)),
(0, 0, 2): ((i + 1, j, VALUE_A), (i, j, VALUE_B), (i, j + 1, VALUE_B)),
(0, 0, 3): ((i, j, VALUE_A), (i, j, VALUE_B), (i + 1, j, VALUE_B)),
(0, 0, 4): ((i, j + 1, VALUE_A), (i, i, VALUE_B), (i + 1, j, VALUE_B)),
(0, 1, 1): ((i, i, VALUE_A), (i, j - 1, VALUE_B), (i, j + 1, VALUE_B)),
(0, 1, 2): ((i, j, VALUE_A), (i, j - 1, VALUE_B), (i, j + 1, VALUE_B)),
(0, 1, 3): ((i, j - 1, VALUE_A), (i, j, VALUE_B), (i, j, VALUE_B)),
(0, 1, 4): ((i, j + 1, VALUE_A), (i, j, VALUE_B), (i, j, VALUE_B)),
(0, 2, 1): ((i, j, VALUE_A), (i, j - 1, VALUE_B), (i, j + 1, VALUE_B)),
(0, 2, 2): ((i + 1, j, VALUE_A), (i, j - 1, VALUE_B), (i, j + 1, VALUE_B)),
(0, 2, 3): ((i, j - 1, VALUE_A), (i, j, VALUE_B), (i + 1, j, VALUE_B)),
(0, 2, 4): ((i, j + 1, VALUE_A), (i, j, VALUE_B), (i + 1, j, VALUE_B))}
return result_map.get((i, j, actions))
for i, j, action in itertools.product(range(4), range(4), range(1, 5)):
print('%s %s %s' % (i, j, action))
T_value = T(i, j, action)
new_T_value = new_T(i, j, action)
if T_value != new_T_value:
raise AssertionError('Error! \nT: %s \nNew T: %s' % (T_value, new_T_value))
我正在尝试实现值迭代算法。 我有一个网格
grid = [[0, 0, 0, +1],
[0, "W", 0, -1],
[0, 0, 0, 0]]
一个动作列表
actlist = {UP:1, DOWN:2, LEFT:3, RIGHT:4}
还有一个奖励函数
reward = [[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0]]
我写了一个函数 T，它 returns 3 个元组的元组。
def T(i,j,actions):
if(i == 0 and j == 0):
if(actions == UP):
return (i,i,0.8),(i,i,0.1),(i,j+1,0.1)
elif(actions == DOWN):
return (i+1,j,0.8),(i,j,0.1),(i,j+1,0.1)
elif(actions == LEFT):
return (i,j,0.8),(i,j,0.1),(i+1,j,0.1)
elif(actions == RIGHT):
return (i,j+1,0.8),(i,i,0.1),(i+1,j,0.1)
elif (i == 0 and j == 1):
if(actions == UP):
return (i,i,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == DOWN):
return (i,j,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == LEFT):
return (i,j-1,0.8),(i,j,0.1),(i,j,0.1)
elif(actions == RIGHT):
return (i,j+1,0.8),(i,j,0.1),(i,j,0.1)
elif(i == 0 and j == 2):
if(actions == UP):
return (i,j,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == DOWN):
return(i+1,j,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == LEFT):
return (i,j-1,0.8),(i,j,0.1),(i+1,j,0.1)
elif(actions == RIGHT):
return (i,j+1,0.8),(i,j,0.1),(i+1,j,0.1)
elif(i == 0 and j == 3):
if(actions == UP):
return (-1,-1,0.8),(-1,-1,0.1),(-1,-1,0.1)
elif(actions == DOWN):
return (-1,-1,0.8),(-1,-1,0.1),(-1,-1,0.1)
elif(actions == LEFT):
return (-1,-1,0.8),(-1,-1,0.1),(-1,-1,0.1)
elif(actions == RIGHT):
return (-1,-1,0.8),(-1,-1,0.1),(-1,-1,0.1)
# 2nd row
elif (i == 1 and j == 0):
if(actions == UP):
return (i-1,j,0.8),(i,j,0.1),(i,j,0.1)
elif(actions == DOWN):
return (i+1,j,0.8),(i,j,0.1),(i,j,0.1)
elif(actions == LEFT):
return (i,j,0.8),(i-1,j,0.1),(i+1,j,0.1)
elif(actions == RIGHT):
return (i,j,0.8),(i-1,j,0.1),(i+1,j,0.1)
elif(i == 1 and j ==1):
if(actions == UP):
return (i,j,0.8),(i,j,0.1),(i,j,0.1)
elif(actions == DOWN):
return (i,j,0.8),(i,j,0.1),(i,j,0.1)
elif(actions == LEFT):
return (i,j,0.8),(i,j,0.1),(i,j,0.1)
elif(actions == RIGHT):
return (i,j,0.8),(i,j,0.1),(i,j,0.1)
elif (i == 1 and j == 2):
if(actions == UP):
return (i-1,j,0.8),(i,j,0.1),(i,j+1,0.1)
elif(actions == DOWN):
return (i+1,j,0.8),(i,j,0.1),(i,j+1,0.1)
elif(actions == LEFT):
return (i,j,0.8),(i-1,j,0.1),(i+1,j,0.1)
elif(actions == RIGHT):
return (i,j+1,0.8),(i-1,j,0.1),(i+1,j,0.1)
elif(i == 1 and j == 3):
if(actions == UP):
return (-2,-2,0.8),(-2,-2,0.1),(-2,-2,0.1)
elif(actions == DOWN):
return (-2,-2,0.8),(-2,-2,0.1),(-2,-2,0.1)
elif(actions == LEFT):
return (-2,-2,0.8),(-2,-2,0.1),(-2,-2,0.1)
elif(actions == RIGHT):
return (-2,-2,0.8),(-2,-2,0.1),(-2,-2,0.1)
# 3rd row
elif(i == 2 and j == 0):
if(actions == UP):
return (i-1,j,0.8),(i,j,0.1),(i,j+1,0.1)
elif(actions == DOWN):
return (i,j,0.8),(i,j,0.1),(i,j+1,1,0.1)
elif(actions == LEFT):
return (i,j,0.8),(i-1,j,0.1),(i,j,0.1)
elif(actions == RIGHT):
return (i,j+1,0.8),(i-1,j,0.1),(i,j,0.1)
elif (i == 2 and j == 1):
if(actions == UP):
return (i,j,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == DOWN):
return (i,j,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == LEFT):
return (i,j-1,0.8),(i,j,0.1),(i,j,0.1)
elif(actions == RIGHT):
return (i,j+1,0.8),(i,j,0.1),(i,j,0.1)
elif(i == 2 and j == 2):
if(actions == UP):
return (i-1,j,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == DOWN):
return (i,j,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == LEFT):
return (i,j-1,0.8),(i-1,j,0.1),(i,j,1)
elif(actions == RIGHT):
return (i,j+1,0.8),(i-1,j,0.1),(i,j,0.1)
elif(i == 2 and j == 3):
if(actions == UP):
return (i-1,j,0.8),(i,j-1,0.1),(i,j,0.1)
elif(actions == DOWN):
return (i,j,0.8),(i,j-1,0.1),(i,j,0.1)
elif(actions == LEFT):
return (i,j-1,0.8),(i-1,j,0.1),(i,j,0.1)
elif(actions == RIGHT):
return (i,j,0.8),(i-1,j,0.1),(i,j,0.1)
这个函数在值迭代函数中被调用：
def value_iteration():
U1 = [[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0]]
while True:
U=U1.copy()
delta = 0
for i in range(len(grid)):
for j in range(len(grid[i])):
U1[i][j] = max(sum(p*(R(k,l)+gamma*U[k][l]) for (k,l,p) in T(i,j,a)) for a in actlist)
print(i,j,U1[i][j])
delta = max(delta, abs(U1[i][j] - U[i][j]))
if delta <= epsilon*(1 - gamma)/gamma:
return U
问题是，for 循环的前两次迭代与输出相得益彰
0 0
0 1
0 2
0 3
1 0
1 1
1 2
1 3
但随后代码因错误停止
ValueError: too many values to unpack (expected 3)
看看** **中的元组，也许就是这个原因
# 3rd row
elif(i == 2 and j == 0):
if(actions == UP):
return (i-1,j,0.8),(i,j,0.1),(i,j+1,0.1)
elif(actions == DOWN):
return (i,j,0.8),(i,j,0.1),**(i,j+1,1,0.1)**
elif(actions == LEFT):
return (i,j,0.8),(i-1,j,0.1),(i,j,0.1)
elif(actions == RIGHT):
return (i,j+1,0.8),(i-1,j,0.1),(i,j,0.1)
elif (i == 2 and j == 1):
if(actions == UP):
return (i,j,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == DOWN):
return (i,j,0.8),(i,j-1,0.1),(i,j+1,0.1)
elif(actions == LEFT):
return (i,j-1,0.8),(i,j,0.1),(i,j,0.1)
elif(actions == RIGHT):
return (i,j+1,0.8),(i,j,0.1),(i,j,0.1)
正如@EdwardMinnix 提到的，您应该使用 map 而永远不要进行这种 if/else 构造。或者，如果有任何具有这些值的模式，请查看 Stragety Pattern.
这会对你有所帮助：
...
VALUE_A = 0.8
VALUE_B = 0.1
def new_T(i, j, actions):
result_map = {(0, 0, 1): ((i, i, VALUE_A), (i, i, VALUE_B), (i, j + 1, VALUE_B)),
(0, 0, 2): ((i + 1, j, VALUE_A), (i, j, VALUE_B), (i, j + 1, VALUE_B)),
(0, 0, 3): ((i, j, VALUE_A), (i, j, VALUE_B), (i + 1, j, VALUE_B)),
(0, 0, 4): ((i, j + 1, VALUE_A), (i, i, VALUE_B), (i + 1, j, VALUE_B)),
(0, 1, 1): ((i, i, VALUE_A), (i, j - 1, VALUE_B), (i, j + 1, VALUE_B)),
(0, 1, 2): ((i, j, VALUE_A), (i, j - 1, VALUE_B), (i, j + 1, VALUE_B)),
(0, 1, 3): ((i, j - 1, VALUE_A), (i, j, VALUE_B), (i, j, VALUE_B)),
(0, 1, 4): ((i, j + 1, VALUE_A), (i, j, VALUE_B), (i, j, VALUE_B)),
(0, 2, 1): ((i, j, VALUE_A), (i, j - 1, VALUE_B), (i, j + 1, VALUE_B)),
(0, 2, 2): ((i + 1, j, VALUE_A), (i, j - 1, VALUE_B), (i, j + 1, VALUE_B)),
(0, 2, 3): ((i, j - 1, VALUE_A), (i, j, VALUE_B), (i + 1, j, VALUE_B)),
(0, 2, 4): ((i, j + 1, VALUE_A), (i, j, VALUE_B), (i + 1, j, VALUE_B))}
return result_map.get((i, j, actions))
for i, j, action in itertools.product(range(4), range(4), range(1, 5)):
print('%s %s %s' % (i, j, action))
T_value = T(i, j, action)
new_T_value = new_T(i, j, action)
if T_value != new_T_value:
raise AssertionError('Error! \nT: %s \nNew T: %s' % (T_value, new_T_value))